Find $S = a + b$ such that for $\forall m \in \left[a\sqrt{\frac{15}{7}} + b\sqrt{\frac{7}{15}}; 2\right)$ then $2x^2 + 2x - mf(x) + 5 = 0$ has root 
Let $f(x)$ be continuos on $\mathbb R$ satisfy $f(0) = 2\sqrt{2}$ and $f(x) > 0, \forall x \in \mathbb R$ and $f(x) f'(x) = (2x+1)\sqrt{1+f^2(x)}$.
For all $m \in \left[a\sqrt{\dfrac{15}{7}} + b\sqrt{\dfrac{7}{15}}; 2\right)$, then $2x^2 + 2x - mf(x) + 5 = 0$ has at least a root. Find $S = a + b$


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*Here is what I did so far

$$f(x) f'(x) = (2x+1)\sqrt{1+f^2(x)}$$
$$\implies \dfrac{f(x) f'(x)}{\sqrt{1+f^2(x)}} = 2x+1$$
$$\implies \int \dfrac{f(x) f'(x)}{\sqrt{1+f^2(x)}} \,dx = \int2x+1 \,dx = x^2 + x + C$$
Let $t = f^2(x) + 1 \implies dt = 2f'(x)f(x) dx$. Therefore:
$$\int \dfrac{1}{2\sqrt{t}} \,dx = x^2 + x + C$$
$$\implies \sqrt{t} = x^2 + x + C$$
$$\implies \sqrt{1 + f^2(x)} = x^2 + x + C \, (2)$$
Since $f(0) = 2\sqrt{2}$, plug in $(2)$ we get $C = 3 \implies f(x) = \sqrt{(x^2+x+3)^2 - 1}$
I can't proceed further from here
 A: The question is equivalent to asking when $$f(x) = (4-m^2) x^4 +(8-2m^2)x^3 +(24-7m^2) x^2 +(20-6m^2) x+25 - 8m^2 $$ has a root.
Let’s consider
$$f’(x) = 4(4-m^2)x^3 +6(4-m^2)x^2 +2(24-7m^2) x+(20-6m^2) \\ = -2 (1 + 2 x) (-10 + 3 m^2 + (m^2 -4)x + (m^2 -4)x^2) $$
So, there is always a turning point at $x=-0.5$.
$$f(-0.5) =\frac{81}{4}-\frac{105m^2}{16} \le 0 \iff m\ge \sqrt{\frac{108}{35}} $$
Now, the leftover quadratic has discriminant $$-11m^4+80m^2-144 \ge 0 \iff \frac{6}{\sqrt{11}} \le m \le 2 $$
You can check that $\sqrt{\frac{108}{35}} \lt \frac{6}{\sqrt{11}} $, and so there’s no need to consider the case where $f$ has three turning points, as $f$ will already have atleast one root for $m\ge \sqrt{\frac{108}{35}} $. Now it only remains to write this square root in the form in the question. For that, manipulate $$a\sqrt {\frac{15}{7}} +b\sqrt{\frac{7}{15}}= \frac{15a+7b}{\sqrt{105}} \overset{\text{must be}}= \sqrt{\frac{108}{35}} \\ \implies \color{magenta}{15a+7b = 18 } $$
Unless you specify more conditions, $a+b$ can take infinitely many values.
A: COMMENT.-Once find out $f(x)=\sqrt{(x^2+x+3)^2-1}$ the function $g(x)=2x^2+2x-mf(x)+5$ has derivative $g'(x)=(2x+1)\left(2-\dfrac{m(x^2+x+3)}{f(x)}\right)$ so take an extremum at $x=-\dfrac12=-0.5$ and $g(-0.5)=4.5-m\dfrac{\sqrt{105}}{16}$
It follows that $g(x)$ has at least a (real) root in a little interval for $m$ whose first element is given by $g(-0.5)=4.5-m\dfrac{\sqrt{105}}{16}=0\iff m=\dfrac{18}{\sqrt{105}}\approx1.756620$ and it is bounded by $2$.
(for example for $m=1.99$ one has $g(9.31)=0$ but for $m\ge2$ there are not (real) root).
Then we have $\dfrac{18}{\sqrt{105}}=a\sqrt{\dfrac{15}{7}}+b\sqrt{\dfrac{7}{15}}=\dfrac{15a+7b}{\sqrt{105}}\Rightarrow \boxed {15a+7b=18}$
Can someone finish?
